Respuesta :
Answer: (a) Percentage of 25 year old men that are above 6 feet 2 inches is 11.5%.
(b) Percentage of 25 year old men in the 6 footer club that are above 6 feet 5 inches are 2.4%.
Step-by-step explanation:
Given that,
Height (in inches) of a 25 year old man is a normal random variable with mean [tex]g=71[/tex] and variance [tex]o^{2} =6.25[/tex].
To find: (a) What percentage of 25 year old men are 6 feet, 2 inches tall
(b) What percentage of 25 year old men in the 6 footer club are over 6 feet. 5 inches.
Now,
(a) To calculate the percentage of men, we have to calculate the probability
P[Height of a 25 year old man is over 6 feet 2 inches]= P[X>[tex]74in[/tex]]
P[X>74] = P[[tex]\frac{X-g}{o}[/tex] > [tex]\frac{74-71}{2.5}[/tex]]
= P[Z > 1.2]
= 1 - P[Z ≤ 1.2]
= 1 - Ф (1.2)
= 1 - 0.8849
= 0.1151
Thus, percentage of 25 year old men that are above 6 feet 2 inches is 11.5%.
(b) P[Height of 25 year old man is above 6 feet 5 inches gives that he is above 6 feet] = P[X, [tex]6ft[/tex] [tex]5in[/tex] - X, [tex]6ft[/tex]]
P[X > [tex]6ft[/tex] [tex]5in[/tex] I X > [tex]6ft[/tex]] = P[X > 77 I X > 72]
= [tex]\frac{P[X > 77]}{P[ X > 72]}[/tex]
= [tex]\frac{P[\frac{X - g}{o}>\frac{77-71}{2.5}] }{P[\frac{X-g}{o} >\frac{72-71}{2.5}] }[/tex]
= [tex]\frac{P[Z >2.4]}{P[Z>0.4]}[/tex]
= [tex]\frac{1-P[Z\leq2.4] }{1-P[Z\leq0.4] }[/tex]
= [tex]\frac{1-0.9918}{1-0.6554}[/tex]
= [tex]\frac{0.0082}{0.3446}[/tex]
= 0.024
Thus, Percentage of 25 year old men in the 6 footer club that are above 6 feet 5 inches are 2.4%.
